The outer surface of a 3×3 Rubik's cube is painted red. You break the Rubik's cube into 27 smaller cubes. You select 1 at random and put it on the table. You see that the 5 visible spaces have no red painting. What is the probability that the bottom face is painted red?
So I am given that the selected cube has 5 faces that are not painted red. This eliminates 20 of the cubes. The remaining cubes are the 6 centered cubes on each of the 6 faces, and the very center cube.
So isn't the probability simply 6/7? There is a hint that says to use Bayes theorem, but I don't see how that comes in. All that is needed seems to be just the definition of conditional probability. Specifically we want $P(A\mid B)$ where $A$ is the event that the selected cube has 1 painted face and $B$ is the event that the selected cube has 5 un-painted faces.
Best Answer
If it’s one of the cubes in the centres of the faces, there is only one chance in six that the red face is on the bottom, but if it’s the central cube, any of the six faces could be on the bottom. To put it a bit differently, the $7$ cubes have a total of $42$ faces, all of which are equally likely to be on the bottom. If you don’t see a red face, you know that the face on the bottom is either one of the $6$ faces of the central cube or the one red face of one of the other $6$ cubes. Thus, it’s one of $12$ faces, and your sample space actually has size $12$. In $6$ of those $12$ cases the bottom face is red, so the probability is actually $\frac12$.